Aspire Faculty ID #15272 · Topic: JAMIA MCA 2016 · Just now
JAMIA MCA 2016

If $z$ be a complex number and $\bar{z}$ be its conjugate, then the number of solutions of $z^{2} + 2\bar{z} = 0$ is …

Solution

Let $z = x + iy$, then $\bar{z} = x - iy$. Substitute: $(x + iy)^{2} + 2(x - iy) = 0$. $\Rightarrow x^{2} - y^{2} + 2ixy + 2x - 2iy = 0$. Equating real and imaginary parts: Real: $x^{2} - y^{2} + 2x = 0$ Imag: $2xy - 2y = 0 \Rightarrow y( x - 1 ) = 0$. If $y=0$, then $x^{2} + 2x = 0 \Rightarrow x = 0, -2$. If $x=1$, then $1 - y^{2} + 2 = 0 \Rightarrow y^{2} = 3$. Hence, total 4 solutions.

Previous 10 Questions — JAMIA MCA 2016

Nearest first

Next 10 Questions — JAMIA MCA 2016

Ascending by ID
Ask Your Question or Put Your Review.

loading...